Question

You need to prepare an acetate buffer of pH 6.50 from a 0.770 M acetic acid solution and a 2.85 M KOH solution. If you have 675 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 6.50? The pKa of acetic acid is 4.76.

Answer 1

Write down the acid dissociation of acetic acid, HAc as

HAc (aq) <======> H⁺ (aq) + Ac^- (aq)

The pH is given by the Henderson-Hasslebach equation as

pH = pK_a + log [Ac^-]/[HAc]

Plug in pH = 6.50 and pK_a = 4.76 to obtain the ration of Ac^-/HAc as

6.50 = 4.76 + log [Ac^-]/[HAc]

====> 1.74 = log [Ac^-]/[HAc]

====> [Ac^-]/[HAc] = antilog (1.74) = 54.954

====> [Ac^-] = 54.954*[HAc] ……..(1)

Again, we are given that [HAc] = 0.770 M; therefore, [Ac^-] = 54.954*(0.770 M) = 42.314 M.

We must have the [Ac^-] in our buffer as 42.314 M. We do not have Ac^- supplied to us, instead, we have HAc and KOH given; offcourse then, an acid-base neutralization reaction occurs to produce Ac^-

Moles HAc present = (volume in L)*(concentration in moles/L) = (675 mL)*(1 L/1000 mL)*(0.770 mole/L) = 0.51975 mole.

Write down the neutralization reaction:

HAc (aq) + KOH (aq) --------> KAc (aq) + H₂O

We can clearly realize that there was no Ac^- present initially and Ac^- is formed as KOH is added to the solution. Therefore, we can write moles KOH added = moles Ac^- formed.

[Ac^-] = (moles Ac^- formed)/(total volume of the solution in L)

Let x mL of 2.85 M KOH be added; therefore, moles KOH added = (volume of KOH added in L)*( concentration of KOH added in moles/L) = (x mL)*(1 L/1000 mL)*(2.85 moles/L) = 0.00285x mole = moles Ac^- formed.

Total volume of the solution formed in L = [(675 + x)mL]*(1 L/1000 mL).

Therefore, 42.314 mole/L = (0.00285 mole)/[(675 + x)mL]*(1000 mL/1 L) = 2.85x/(675 + x) mole/L

=====> 42.314*(675 + x) = 0.00285x

=====> 28561.95 + 42.314x = 2.85x

Ignore negative sign to obtain,

39.464x = 28561.95

=====> x = 723.70

Therefore, we need to add 723.70 mL KOH (ans).

The answer definitely looks wrong. The reason is that you want to prepare a buffer solution of pH 6.50 starting from an acid whose pK_a is 4.76. For ideal buffer action, the pH of the buffer must lie within 1 unit on either side of the pK_a. In this case, the pH must lie within 3.76-5.76; offcourse the pH is almost 2 units higher than the pK_a and hence the calculation gives erroneous results. Please check the pH. If it is 4.50, then the calculations will nicely fall in place (ans).